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Math Help - Another Tricky Integral

  1. #1
    Member RedBarchetta's Avatar
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    Another Tricky Integral

    <br />
\int {\frac{1}<br />
{{1 + \cos x}}dx = \tan \tfrac{x}<br />
{2} + C} <br />

    Alright. Would I multiply the bottom part by [1-cos(x)]?

    Here's what I tried:

    <br />
\int {\frac{{1(1 - \cos x)}}<br />
{{(1 + \cos x)(1 - \cos x)}}dx}  = \int {\frac{{1 - \cos x}}<br />
{{1 - \cos ^2 x}}dx} <br />

    Now I split these up into two integrals:

    <br />
\begin{gathered}<br />
  \sin ^2 x + \cos ^2 x = 1 \Rightarrow \sin ^2 x = 1 - \cos ^2 x \hfill \\<br />
  \int {\frac{1}<br />
{{\sin ^2 x}}dx}  - \int {\frac{{\cos x}}<br />
{{\sin ^2 x}}dx}  \hfill \\<br />
  \int {\csc ^2 xdx}  - \int {\cot x\csc xdx}  = \csc x - \cot x + c \hfill \\ <br />
\end{gathered} <br />

    ....am I right?.....or where did I go wrong.

    Thank you.
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  2. #2
    o_O
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    It can be shown that: \tan \left(\frac{x}{2}\right) = \frac{1 - \cos x}{\sin x} = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \csc x - \cot x

    Basically, yeah you're good
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  3. #3
    Moo
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    Hello,
    Quote Originally Posted by RedBarchetta View Post
    <br />
\int {\frac{1}<br />
{{1 + \cos x}}dx = \tan \tfrac{x}<br />
{2} + C} <br />

    Alright. Would I multiply the bottom part by [1-cos(x)]?


    Here's what I tried:

    <br />
\int {\frac{{1(1 - \cos x)}}<br />
{{(1 + \cos x)(1 - \cos x)}}dx}  = \int {\frac{{1 - \cos x}}<br />
{{1 - \cos ^2 x}}dx} <br />

    Now I split these up into two integrals:

    <br />
\begin{gathered}<br />
  \sin ^2 x + \cos ^2 x = 1 \Rightarrow \sin ^2 x = 1 - \cos ^2 x \hfill \\<br />
  \int {\frac{1}<br />
{{\sin ^2 x}}dx}  - \int {\frac{{\cos x}}<br />
{{\sin ^2 x}}dx}  \hfill \\<br />
  \int {\csc ^2 xdx}  - \int {\cot x\csc xdx}  = \csc x - \cot x + c \hfill \\ <br />
\end{gathered} <br />

    ....am I right?.....or where did I go wrong.

    Thank you.
    Bravo ! Your answer is correct.

    if you want to transform it into tan(x/2) :

    \frac 1{\sin(x)}-\frac{\cos(x)}{\sin(x)}=\frac{1-\cos(x)}{\sin(x)}=\frac{1-(1-2\sin^2 \tfrac x2)}{2 \cos \tfrac x2 \sin \tfrac x2}

    =\frac{2 \sin^2 \tfrac x2}{2 \cos \tfrac x2 \sin \tfrac x2}

    Is there anything you can do ?
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  4. #4
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    <br />
\int {\frac{1}<br />
{{1 + \cos x}}dx = \tan \tfrac{x}<br />
{2} + C} <br />

    Alright. Would I multiply the bottom part by [1-cos(x)]?

    Here's what I tried:

    <br />
\int {\frac{{1(1 - \cos x)}}<br />
{{(1 + \cos x)(1 - \cos x)}}dx} = \int {\frac{{1 - \cos x}}<br />
{{1 - \cos ^2 x}}dx} <br />

    Now I split these up into two integrals:

    <br />
\begin{gathered}<br />
\sin ^2 x + \cos ^2 x = 1 \Rightarrow \sin ^2 x = 1 - \cos ^2 x \hfill \\<br />
\int {\frac{1}<br />
{{\sin ^2 x}}dx} - \int {\frac{{\cos x}}<br />
{{\sin ^2 x}}dx} \hfill \\<br />
\int {\csc ^2 xdx} - \int {\cot x\csc xdx} = \csc x - \cot x + c \hfill \\ <br />
\end{gathered} <br />

    ....am I right?.....or where did I go wrong.

    Thank you.
    The easier way:

    1 + \cos x = 1 + \cos^2 \left( \frac{x}{2} \right) - \sin^2 \left( \frac{x}{2} \right)

    using the double angle formula

     = 2 \cos^2 \left( \frac{x}{2} \right).

    Therefore your integral becomes \int \frac{1}{2 \cos^2 \left( \frac{x}{2} \right)} \, dx = \frac{1}{2} \int \sec^2 \left( \frac{x}{2} \right) \, dx = \tan \left( \frac{x}{2} \right) + C.
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